4
$\begingroup$

Given two path algebras $A$ and $B$, for example, A=: $1\to2$ B=: $1\to2\to3$, is the tensor product of A and B over a field $k$ a path algebra? if yes, how to represent it by a quiver? also, how to construct $(A,B)$-bimodules?

Improve this question
$\endgroup$
1
  • 8
    $\begingroup$ 1) Paths algebras (over non-trivial quivers) have homological dimension $1$, while the tensor product of two path algebras has homological dimension $\geq 2$, so it is not a path algebra 2) What kind of question is this? There are free bimodules and any other bimodule is a quotient of a free one. $\endgroup$
    – Fernando Muro
    May 9, 2012 at 10:25

2 Answers 2

Reset to default
1
$\begingroup$

It can be realized as a quotient of the Cartesian product of path algebras in question. You have to mod out by some commutativity relations.

Improve this answer
$\endgroup$
3
  • 1
    $\begingroup$ How does that work? Specifically, which element in the tensor product would be the image of the element $(a,b)$ of the Cartesian product? The "obvious" choice, $a\otimes b$ won't work, because addition in the Cartesian product ($(a,b)+(c,d)=(a+c,b+d)$) doesn't match addition in the tensor product, so you won't have a homomorphism. (Multiplication by scalars from $k$ doesn't match either unless $k$ is the two-element field.) $\endgroup$
    – Andreas Blass
    May 9, 2012 at 12:51
  • $\begingroup$ I think he means take the Cartesian product of the quivers and then put in commutativity relations. For instance, if you take the quiver 1->2 and take the tensor product of this path algebra with itself you get the incidence algebra of the diamond 1 / \ a b \ / 0 The Hasse diagram is the quiver of this algebra, but you have the relation that the two paths to the top are the same. $\endgroup$
    – Benjamin Steinberg
    May 9, 2012 at 21:12
  • $\begingroup$ Hmm, my picture of the Hasse diagram of a diamond didn't draw well in the comments. $\endgroup$
    – Benjamin Steinberg
    May 9, 2012 at 21:13
1
$\begingroup$

What you want is

here.

Basically you take the 1-skeleton of the products of the quivers and you say that certain paths e.g. (edge,vertex)(vertex,edge)=(vertex,edge)(edge,vertex).

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.